what i mean is that when say, an artillery shell is fired, or a piece of debris from an explosion is launched, it goes up and away at an almost 45 degree angle until its upwards momentum is no longer sufficient to fight gravity before dropping pretty much straight down. second, given the angle of impact on the ground, could you calculate backwards from there the initial velocity and launch site of the projectile?
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You answered your question completely.
Gravity pulls on all objects the same.
Step 1: load the shell. The shell is in the gun and the bolt is locked. The shell isn’t moving. The shell IS accelerating toward the center of the earth at 9.8 meters per second, every second (9.8 m/s/s, or 9.8 m/s²).
Step 2: bang! The shell is accelerated by the charges like 1 million meters per second for a tiny fraction of a second. It reaches it’s highest speed right at the end of the barrel.
The moment it leaves the barrel it starts slowing down because gravity is pulling it down at 9.8 m/s².
Eventually, gravity steals enough energy that the bullet can’t continue going up, gravity takes over, and the shell comes down.
The math you’re talking about is also used during aiming. They know the distance to the target. They can set how hard they shoot the shell, but only in increments.
Have you seen artillery loading? The shell goes in first then they load bags of explosives. Each bag equals the same projectile force and you can load up to 5. Loading only 2 means the shell can’t go as far as loading 5.
So they know how far they need to shoot, how heavy the shell is, and how far each bag will get them. The only thing left to determine is the launch angle.
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